Math inequality proof If $a, b$ are positive real numbers and $a + b = 1$, prove that
$$
\left(a +\frac{1}{a}\right)^2 + \left(b +\frac{1}{b}\right)^2 \geq \frac{25}{2}
$$
Thank you.
 A: yes,use $Cuachy-Schwarz$ inequality,we have 
$$\left[\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\right][1+1]\ge\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2$$
use $Cauchy-Schwarz $ inequality,we have
$$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)(a+b)\ge(1+1)^2$$
A: Replace $b$  by $(1-a)$. Now, compute the derivative of the function; it has only one positive real root at $a=\frac12$. At this point, the second derivative is positive, then $a=\frac12$ corresponds to a minimum and, at this point, the function value is $\frac{25}{2}$
A: Let $\displaystyle \phi(x) = \left(x + \frac{1}{x}\right)^2$. Since
$$\frac{d^2}{dx^2} \phi(x) = 2 + \frac{6}{x^4} > 0,$$
$\phi(x)$ is a strictly convex function for $x > 0$. 
By Jensen's inequality, we have
$$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 = \phi(a)+\phi(b) \ge 2 \phi(\frac{a+b}{2}) = 2 \left(\frac12 + 2\right)^2 = \frac{25}{2}$$
A: It's easily seen that, for $a+b=1$ with $a,b\gt0$, the expression
$$ab+{1\over ab}$$
is minimized when $a=b=1/2$.  (In general, $x+{1\over x}$ is smallest when $x$ is as close to $1$ as possible.  If $a+b=1$, the closest $ab$ gets to $1$ is when $a=b=1/2$.)  It follows that
$$\begin{align}
\left(a+{1\over a} \right)^2+\left(b+{1\over b} \right)^2&=a^2+b^2+{1\over a^2}+{1\over b^2}+4\cr
&=(a-b)^2+\left({1\over a}-{1\over b} \right)^2+2\left(ab+{1\over ab}\right)+4\cr
&\ge 2\left(ab+{1\over ab}\right)+4\cr
&\ge 2\left({1\over4}+{1\over1/4}\right)+4\cr
&={25\over2}
\end{align}$$
